Answer
Minimum: $f(0,1.211) \approx -1.403$, $f(0,-1.273) \approx -3.890$
Saddle point at $(0,0.347)$
Lowest point: $(0,-1.273, -3.890)$
Work Step by Step
Second derivative test: Some noteworthy points to calculate the local minimum, local maximum and saddle point of $f$.
1. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum.
2.If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum.
3. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is not a local minimum and local maximum or, a saddle point.
Critical points are: $(0,1.211),(0,-1.273),(0,0.347),(0,-1.273)$
For $(x,y)=(0,-1.273)$
$D(0,-1.273)=67.8964 \gt 0$ and $f_{xx}=2 \gt 0$
For $(x,y)=(0,1.211)$
$D(0,1.211) =54.467 \gt 0$ and $f_{xx}=2 \gt 0$
Thus, when $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum.
For $(x,y)=(\pm 0.720,0.259)$
$D(0,0.347) =-8.8550 \lt 0$ ; saddle point.
Hence,
Minimum: $f(0,1.211) \approx -1.403$, $f(0,-1.273) \approx -3.890$
Saddle point at $(0,0.347)$
Lowest point: $(0,-1.273, -3.890)$
